LightOJ 1102 Problem Makes Problem（卢卡斯+阶乘处理+隔板法）

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Description

As I am fond of making easier problems, I discovered a problem. Actually, the problem is 'how can you make n by adding k non-negative integers?' I think a small example will make things clear. Suppose n=4 and k=3. There are 15solutions. They are

1.      0 0 4

2.      0 1 3

3.      0 2 2

4.      0 3 1

5.      0 4 0

6.      1 0 3

7.      1 1 2

8.      1 2 1

9.      1 3 0

10.    2 0 2

11.    2 1 1

12.    2 2 0

13.    3 0 1

14.    3 1 0

15.    4 0 0

As I have already told you that I use to make problems easier, so, you don't have to find the actual result. You should report the result modulo 1000,000,007.

Input

Input starts with an integer T (≤ 25000), denoting the number of test cases.

Each case contains two integer n (0 ≤ n ≤ 106) and k (1 ≤ k ≤ 106).

Output

For each case, print the case number and the result modulo 1000000007.

Sample Input

4

4 3

3 5

1000 3

1000 5

Sample Output

Case 1: 15

Case 2: 35

Case 3: 501501

Case 4: 84793457
 

题意 ： 输入n,k，输出 将n分成k个非负整数之和的方案数。

分析：隔板法的经典应有，具体原理：隔板法详解：​​点这里​​

ans = C(n+k-1,k-1);

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
#define LL __int64
const int N=2000100;
const int p=1000000007;

LL fac[N];
 
void init()   //预处理阶乘
{
    LL i;
    fac[0] =1;
    for(i =1; i < N; i++)
        fac[i] = fac[i-1]*i % p;
}
 
LL exp_mod(LL a, LL b)
{
    LL tmp = a % p, ans =1;
    while(b)
    {
        if(b & 1)  ans = ans * tmp % p;
        tmp = tmp*tmp % p;
        b >>=1;
    }
    return  ans;
}
 
LL C(LL n, LL m)
{
    if(m > n)
        return 0;
    return  fac[n]*exp_mod(fac[m]*fac[n-m], p-2) % p;//逆元
}
 
LL Lucas(LL n, LL m)
{
    if(m ==0)
        return 1;
    return  (C(n%p, m%p)*Lucas(n/p, m/p))%p;
}
 
 
int main()
{
    int T;
    LL n,m,cas=1;
    
    init();
    
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld %lld",&n,&m);
        printf("Case %lld: %lld\n",cas++,Lucas(n+m-1,m-1));
    }
    return 0;
}